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APM346-2015S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 14, 2015, 07:47:02 PM

Title: FE Problem 5
Post by: Victor Ivrii on April 14, 2015, 07:47:02 PM
Consider Laplace equation in the half-strip
u_{xx} +u_{yy}=0 \qquad  x>0, \ 0 <y< \pi
with the boundary conditions
u (x,0)=0,\qquad
with $g(x)=1$   and condition $\max |u|<\infty$.

a Write the associated eigenvalue problem.

b  Find all  eigenvalues and corresponding eigenfunctions.

c  Write the solution in the form of  a series expansion.
Title: Re: FE Problem 5
Post by: Zacharie Leger on April 15, 2015, 07:22:27 PM
Hopeful solution
a. The eigenvalue problem can be found using separation of variables, hence if we let $u(x,y)=X(x)Y(y)$ and plug into the Laplace equation we can get
$$\frac{X''(x)}{X(x)}+\frac{Y''(y)}{Y(y)}=0 \Rightarrow -\frac{X''(x)}{X(x)}=\frac{Y''(y)}{Y(y)}=-\lambda$$

b. The boundry conditions the problem tell us that $\lambda >0$, since we have periodic boundary conditions for $Y(y)$. Furthermore, if we let $\lambda =\omega ^2,\,\omega >0$ we get the following solution
$$Y(y)=A\sin(\omega y)+B\cos(\omega y)$$
Applying the boundry conditions on the solution we see that
$$Y(0)=A\sin(\omega (0))+B\cos(\omega (0))=B=0\,\mathrm{and}\, Y(\pi)=A\sin(\omega (\pi))=0\Rightarrow \omega =n$$
$$\Rightarrow Y_n (y)=A_n \sin(ny)$$
as for the $X(x)$ solution we get
Applying the boundry conditions all we can say is that
$$X_n (x)=C_n\cosh(nx)+\sinh(nx)$$
The full solution can be expressed as
$$u(x,y)=\sum\limits_{n=1}^\infty [(C_n \cosh(nx)+\sinh(nx))A_n\sin(ny)]$$

c. From this we get
$$u_x (0,y)=1=\sum\limits_{n=1}^\infty [(nC_n \sinh(n(0))+n\cosh(n(0)))A_n \sin(ny)]=\sum\limits_{n=1}^\infty [A_n \sin(ny)]$$
The coefficiants are given by
$$A_n=\frac{2}{pi} \int_0^\pi (1)\sin(ny)\mathrm{d}x=\frac{2}{pi}\frac{-cos(nx)}{n}|_0^\pi = \frac{2}{\pi}\frac{1+(-1)^2}{n}$$
Finally, we get
$$u(x,y)=\sum\limits_{n=1}^\infty \frac{2}{\pi}\frac{1+(-1)^2}{n}[(C_n \cosh(nx)+\sinh(nx))\sin(ny)]$$
Title: Re: FE Problem 5
Post by: Victor Ivrii on April 17, 2015, 08:20:44 AM
Looking at $u(x,y)=X(x)Y(y)$ we get $Y''+\lambda Y=0$, $Y(0)=Y(\pi)=0$ and therefore $\lambda_n=n^2$, $Y_n=\sin (ny)$, $n=1,2,\ldots$,
$X_n''-n^2 X_n=0$, $X_n=A_ne^{-nx}+B_n e^{nx}$ with $B_n=0$ (or solution would be unbounded), and
u(x,y)=\sum_{n=1}^\infty A_n e^{-nx}\sin(ny);
u_x(0,y)=\sum_{n=1}^\infty  -n A_n\sin(ny)=1
nA_n =-\frac{2}{\pi} \int_0^\pi \sin (ny)\,dy= \frac{2}{n\pi}\cos(ny)\Bigr|_0^\pi = \left\{\begin{aligned}
&0 && n=2m,\\
-&\frac{4}{n\pi} && n=2m+1
u(x,y)=-\sum_{m=0}^\infty \frac{4}{(2m+1)^2\pi} e^{-(2m+1)x}\sin((2m+1)y);